I've been finding some difficulties in an exercise about the stochastic integral:
Consider the following stochastic process: $$X_t=\int_0^t\sigma_udW_u$$ Where $\sigma_u$ is a cadlag deterministic adapted process, $W_u$ is a standard Brownian motion and $\{\mathcal{F}_t\}_{t\geq 0}$ is the standard augmented filtration. Is it true that $X_t-X_s$ is independent on $\mathcal{F}_s$?
MY ATTEMPT
Note that the process $X_t-X_s=\int_s^t\sigma_udW_u$ and for any partition of the interval $[s,t]$ of the type $\pi^n=\{s=t_0^n<t_1^n<...<t^n_{k_n}=t\}$ with $\max|t^n_i-t^n_{i-1}|\xrightarrow[n\to\infty]{}0$: $$\lim_{n\to\infty}\sum_{i=1}^{k_n}\sigma_{t^n_{i-1}}(W_{t^n_i}-W_{t^n_{i-1}})\xrightarrow[n\to\infty]{\mathbb{P}}\int_s^t\sigma_udW_u$$ We have that for any $n$ the LHS is independent on $\mathcal{F}_s$ thanks to the independece properties of Brownian increments ($\sigma_s$ is deterministic).
Is there any result that allows me to say that also the limit is independent on $\mathcal{F}_s$?
Since the Itô integral is joint Gaussian it suffices to show that for $X_{t}=\int^t \sigma_r dW_{r}$
$$E[X_{s}(X_{t}-X_{s})]=0.$$
A càdlàg function is bounded on $[0,T]$ How to prove that càdlàg (RCLL) functions on $[0,1]$ are bounded?, and so you can just dominated convergence theorem to pass to the limit from the $L^2$-approximation with the dyadics partitions
$$E[X_{s}(X_{t}-X_{s})]=\lim_{n\to +\infty}E[X_{s}^{n}(X^{n}_{t}-X^{n}_{s})]=0.$$
Joint Gaussian
For $t^{n}_{k_{n}}=t^{n}_{k_{n}}(t)$ dyadic-approximation to $t>0$, the sums
$$\left(\sum_{i=1}^{k_{n}(t)}\sigma_{t^n_{i-1}}(W_{t^n_i}-W_{t^n_{i-1}})\right)_{t> 0}$$
form a joint-Gaussian process since sum of independent Gaussians is Gaussian eg. see here
Then as mentioned here The ito integral is gaussian, we simply take limits of their characteristic functions to get the Itô-integrals also having joint Gaussian.